That is, we take the difference between the integral of the true path and the integral of the variated path, where S is the action defined by $\int_{t_1}^{t_2} L(x,\dot{x},t) dt$ and $\delta{S}$ is the variation in action (in the graph it is the distance between the blue line and the red line at any t value). Where do I proceed from there? I don't understand this line in the derivation:

Namely, what is the function O? Where are the partial derivatives coming from? The only thing I know to do is to combine the two integrals because the times and the integration variables are the same. How do I get all the mess to the left of $L(x,\dot{x},t)$ in the equation above?

3 Answers
3

The partial derivatives appear as a result of the chain rule for partial derivatives; this takes place implicitly when the variation is applied to the Lagrangian functional. The big O notation says that there are higher order terms which can be ignored in the variational limit.

The web page shows a couple of ways to obtain the result. Euler used purely geometric methods, and Lagrange invented the variational technique with the delta process.

$\begingroup$I meant to say that though they are differentially related, they are linearly independent and on that base calculus of variation is built. Any takers?$\endgroup$
– NarasimhamJan 18 at 15:07

Namely, what is the function O? Where are the partial derivatives coming from? The only thing I know to do is to combine the two integrals because the times and the integration variables are the same. How do I get all the mess to the left of L(x,x˙,t) in the equation above?

These are good questions. The usual derivation is a heuristic derivation and typically a rigorous derivation of the EL equations in the calculus of variations, at the level of rigour, that a first year undergraduate mathematics degree will approach the calculus of one real variable appears to be fairly rare in the literature.

Abraham & Marsden, in their book, Foundations of mechanics, take a geometric approach, which allows them to deduce the EL equations in a rigorous manner, admitting that an analytic derivation is difficult.

$\begingroup$This is good for intuition, but the EL equations uses functional derivatives.$\endgroup$
– InertialObserverMar 19 at 18:03

$\begingroup$The way I've always seen it, the way to rigorously define these functional derivatives is in terms of ordinary derivatives. I.e. you define a new path as $x(t)$ by $x(t) = x_0(t) + \epsilon \delta x(t)$, where $x_0(t)$ is the path that minimizes the action and $\delta x(t)$ is an arbitrary but fixed function (that vanishes at the end points) and $\epsilon$ is a number. Then you plug in $x(t)$ into Lagrangian, consider the action as a function of $\epsilon$, and take an ordinary derivative with respect to $\epsilon$.$\endgroup$
– AlexMar 19 at 22:39

$\begingroup$No, you're right, but you might want to mention that in your answer.$\endgroup$
– InertialObserverMar 19 at 22:39

$\begingroup$... or equivalently to doing the derivative wrt $\epsilon$: at each value of $t$ in the integrand you expand the integrand to first order in $\epsilon$ and demand that the variation in the action is 0 (at first order in $epsilon$) for every possible perturbation $\delta x(t)$.$\endgroup$
– AlexMar 19 at 22:42